Question

What volume of 0.205 m k3po4 solution is necessary to completely react with 130 ml of 0.0116 m nicl2?

Answer 1

Following chemical reaction is involved for the reaction between K3PO4 and NiCl2

2K3PO4 + 3NiCl2 → 6KCl + Ni3(PO4)2

Volume of solutions is converted to litres for present calculations (1litre = 1000ml)
number of moles of NiCl2 = 0.130 × 0.0116
=0.001508 mol

Now,
`\frac{\text{mole of NiCl2 }}{\text{mole of K3PO4}} = (3)/(2)`
∴ mole of K3PO4 =
`(0.001508X2)/(3)`
=0.001005 mol
given that, K3PO4 is 0.205 M
i.e. 0.205 mol of K3PO4 is present in 1 litre of solution
then 0.001005 mol is present in X litre of solution
∴X =
`(0.001005)/(0.205)` =0.004902 litre =4.902ml